Solution:

For wind instruments, pitch \( f \) is proportional to the speed of sound in air \( v_{\text{air}} propto \sqrt{T_{\text{temp}}} \). As temperature rises, \( v_{\text{air}} \) increases, so pitch rises. For string instruments, \( f propto v_{\text{string}} = \sqrt{T_{\text{tension}}/\mu} \). As temperature rises, the string expands, reducing tension \( T_{\text{tension}} \), so \( v_{\text{string}} \) decreases and pitch falls. Reason R accurately explains this behavior for both cases.